proof of Mollweide’s equations

We transform the equation

$$(a+b)\sin \frac{\gamma }{2}=c\cos \left(\frac{\alpha -\beta }{2}\right)$$

to

$$a\cos \left(\frac{\alpha }{2}+\frac{\beta }{2}\right)+b\cos \left(\frac{\alpha }{2}+\frac{\beta }{2}\right)=c\cos \frac{\alpha }{2}\cos \frac{\beta }{2}+c\sin \frac{\alpha }{2}\sin \frac{\beta }{2},$$

using the fact that $\gamma =\pi -\alpha -\beta$. The left hand side can be further expanded, so that we get:

$$a\left(\cos \frac{\alpha }{2}\cos \frac{\beta }{2}-\sin \frac{\alpha }{2}\sin \frac{\beta }{2}\right)+b\left(\cos \frac{\alpha }{2}\cos \frac{\beta }{2}-\sin \frac{\alpha }{2}\sin \frac{\beta }{2}\right)=c\cos \frac{\alpha }{2}\cos \frac{\beta }{2}+c\sin \frac{\alpha }{2}\sin \frac{\beta }{2}.$$

Collecting terms we get:

$$(a+b-c)\cos \frac{\alpha }{2}\cos \frac{\beta }{2}-(a+b+c)\sin \frac{\alpha }{2}\sin \frac{\beta }{2}=0.$$

Using $s:=\frac{a+b+c}{2}$ and using the equations

$\sin {\displaystyle \frac{\alpha }{2}}$ $=$ $\sqrt{{\displaystyle \frac{(s-b)(s-c)}{bc}}}$

$\cos {\displaystyle \frac{\beta }{2}}$ $=$ $\sqrt{{\displaystyle \frac{s(s-a)}{bc}}}$

we get:

$$2\frac{s(s-c)}{c}\sqrt{\frac{(s-a)(s-b)}{ab}}-2\frac{s(s-c)}{c}\sqrt{\frac{(s-a)(s-b))}{ab}}=0,$$

which is obviously true. So we can prove the first equation by going backwards. The second equation can be proved in quite the same way.

Title	proof of Mollweide’s equations
Canonical name	ProofOfMollweidesEquations
Date of creation	2013-03-22 12:50:10
Last modified on	2013-03-22 12:50:10
Owner	mathwizard (128)
Last modified by	mathwizard (128)
Numerical id	5
Author	mathwizard (128)
Entry type	Proof
Classification	msc 51-00